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Re: If 9^(2x + 5) = 27^(3x − 10), then x = [#permalink]
Bunuel wrote:
If 9^(2x + 5) = 27^(3x − 10), then x =

A. 3
B. 6
C. 8
D. 12
E. 15




The above equation can be written as \(3^{2(2x + 5)} = 3^{3(3x - 10)}\)

since the base are equal we can opt out, therefore we have

2(2x + 5) = 3(3x - 10)

4x + 10 = 9x - 30

x = 8
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If 9^2X+5 [#permalink]
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If \(9^{2x+5} = 27^{3x-10}\), then x=

A. 3

B. 6

C. 8

D. 12

E. 15
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Re: If 9^2X+5 [#permalink]
1
Carcass wrote:
If \(9^{2x+5} = 27^{3x-10}\), then x=

A. 3

B. 6

C. 8

D. 12

E. 15


Rewrite each side with the SAME BASE.

So, we can replace 9 with 3², and replace 27 with 3³ to get: \((3²)^{2x+5} = (3³)^{3x-10}\)

Apply the Power of a Power law to get: \(3^{4x+10} = 3^{9x-30}\)

This means that \(4x+10=9x-30\)

Add 30 to both sides to get: \(4x+40=9x\)

Subtract 4x from both sides to get: \(40=5x\)

Solve: x = 8

Answer: C

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Re: If 9^2X+5 [#permalink]
Carcass wrote:
If \(9^{2x+5} = 27^{3x-10}\), then x=

A. 3

B. 6

C. 8

D. 12

E. 15


3^2(2x + 5) = 3^3(3x - 10)

4x + 10 = 9x - 30

-5x = -40
x = 8
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Re: If 9^2X+5 [#permalink]
thanks
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Re: If 9^(2x + 5) = 27^(3x - 10), then x = [#permalink]
1
9^(2x + 5) = 27^(3x − 10)
3^2(2x + 5) = 3^3 (3x - 10)
2(2x + 5) = 3(3x - 10)
4x + 10 = 9x - 30
40 = 5x
x = 8
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Re: If 9^(2x + 5) = 27^(3x - 10), then x = [#permalink]
c = 8
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Re: If 9^(2x + 5) = 27^(3x 10), then x = [#permalink]
1
Given that \(9^{(2x + 5)}\) = \(27^{(3x − 10)}\) and we need to find the value of x

\(9^{(2x + 5)}\) = \(27^{(3x − 10)}\)
=> \((3^2)^{(2x + 5)}\) = \((3^3)^{(3x − 10)}\)
=> \(3^{2*(2x + 5)}\) = \(3^{3*(3x − 10)}\)
=> \(3^{(4x + 10)}\) = \(3^{(9x − 30)}\)

Since base is same (3) in both left hand side and right hand side so even the power will be the same
=> 4x + 10 = 9x − 30
=> 9x - 4x = 10 + 30 = 40
=> 5x = 40
=> x = \(\frac{40}{5}\) = 8

So, Answer will be C
Hope it helps!
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Re: If 9^(2x + 5) = 27^(3x - 10), then x = [#permalink]
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